5) Let \( R \) denote an integral domain. Let \( W \) denote a finitely generated \( R \)-module that is nonzero
and torsion. Prove that the Annihilator of \( W \) is nonzero.
Proof. Suppose \( W = Rw_1 + Rw_2 + \cdots + Rw_n \), where \( w_i \in W \) and \( r_i w_i = 0 \) for \( 0 \neq r_i \in R \)
(such \( r_i \) exist because \( W \) is torsion). Put \( r = r_1 \cdot r_2 \cdot r_3 \cdots r_n. \) Then \( r \neq 0 \) because its
factors aren't \( 0 \) and \( R \) is an integral domain.
Furthermore, \( r \) annihilates all of \( W. \) For let \( w = s_1 w_1 + s_2 w_2 + \cdots + s_n w_n \) be a
generic element of \( W, \) where \( s_1 \in R. \) Then \( rw = r(s_1 w_1 + s_2 w_2 + \cdots + s_n w_n) = \)
\( (r_1 \cdot r_2 \cdot r_3 \cdots r_n) \cdot (s_1 w_1 + s_2 w_2 + \cdots + s_n w_n) = \)
\( s_1 r_2 \cdots r_n(r_1 w_1) + s_2 r_1 r_3 \cdots r_n(r_2 w_2) + \cdots \). This last is just a rearrangement of scalars to get
\( r_i \) next to \( w_i \), legitimate because \( R \) is a commutative ring. But by hypothesis, each of the parenthesized terms
in the last expression is \( 0 \), so \( rw = 0, \) showing that \( r \) is a nonzero annihilator of \( W. \) QED.
Note: This is part of Ex. §10.3,#5 in Dummit & Foote.